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Write the real displacement as $U_n(t)=\mathrm{Re}[u{e}^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store energy in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing $a$, with longitudinal displacements along the chain.

The two spring forces on site $n$ giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, $\omega \simeq a\sqrt{\alpha /M}|q|$, and the first Brillouin zone ends at $|q|=\pi /a$. The maximum angular frequency is ${\omega }_m=2\sqrt{\alpha /M}$.

Time averaging uses $⟨[\mathrm{Re}(z{e}^{i\omega t}){]}^2⟩=|z{|}^2/2$. ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.

Use periodic boundary conditions on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi )$. Using folded wave-number counting for a monatomic chain combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.

For positive $q$, $\omega ={\omega }_m\sin (qa/2)$ andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

For a small longitudinal disturbance, let the equilibrium pressure and mass density be $p_0,{\rho }_0$, their perturbations be $p^{\prime },{\rho }^{\prime }$, and the fluid velocity be $v$. Linearizing conservation of mass and the inviscid momentum equation in one dimension givesWhen compression is adiabatic, the equation of state gives $p^{\prime }=c_s^2{\rho }^{\prime }$ with $c_s^2=(\partial p/\partial \rho {)}_s$. Differentiating and eliminating $v$ yieldsfor an ideal gas with heat-capacity ratio $\gamma =C_p/C_v$ and molar mass $M_{\mathrm{mol}}$. Compression raises temperature as well as density, explaining the factor $\gamma$ compared with the isothermal value $\sqrt{p_0/{\rho }_0}$.

A rightgoing plane sound wave has pressure and particle velocity in phase, $p^{\prime }={\rho }_0{c}_{s}v$, and mean intensity $⟨p^{\prime }v⟩={\hat{p}}^2/(2{\rho }_0{c}_s)={\rho }_0{c}_s{\hat{v}}^2/2$ for peak amplitudes. Particle displacement is a quarter-cycle out of phase with particle velocity: particles oscillate locally while the compression pattern transports energy. This continuum linear description requires wavelength much larger than the mean free path, small pressure perturbations and negligible viscous attenuation. Adiabatic propagation requires heat exchange across the compressed region to be slow on the oscillation time; strong thermal coupling would instead favour an isothermal response.

To settle quickly after loading, critical damping is preferable to the supplied weak damper:Using the strict-bound mass instead changes this by less than $1\%$. This eliminates free ringing while avoiding the slow tail of severe overdamping. In the stipulated external-force model, greater damping also reduces the steady forced amplitude; there is no finite optimum based solely on that amplitude. A critical or near-critical value is a practical compromise for rapid settling. If the disturbance were specified as floor displacement transmitted through the legs, damping would introduce an additional transmission tradeoff, but that is a different forcing model.

Measure downward displacement relative to the former mean equilibrium. The added $M=20\mathrm{kg}$ changes the equilibrium byAssume the load is deposited without a significant impact impulse and then moves with the table. The moving mass becomes $m+M$. Ignoring the tiny pre-existing $10\mathrm{Hz}$ vibration in the sketch, the added-load step response of a supported table isFor the estimated mass, $\beta =0.194{\mathrm{s}}^{-1}$, the envelope time is $1/\beta =5.15\mathrm{s}$, and the period is about $1.13\mathrm{s}$. The table overshoots and rings about the new equilibrium, with the displacement starting continuously at zero and with zero initial slope. A real impact adds an initial-velocity transient; the problem gives no impact history. The residual driven vibration, of order a micrometre, can be superposed.

The four supporting springs act in parallel, so their total spring constant is $k=4.00\times 10^4\mathrm{N}{\mathrm{m}}^{-1}$. The driving angular frequency is $\omega =2\pi (10)=62.83{\mathrm{s}}^{-1}$. The force-driven vibration isolation estimate givesAt this mass, $f_0=\sqrt{k/m}/(2\pi )=0.894\mathrm{Hz}\ll 10\mathrm{Hz}$, and $Q=\sqrt{mk}/b\simeq 14.2$, so the assumptions are consistent. The exact forced amplitude isIt is indeed close to $1\mu \mathrm{m}$, but it is slightly above the strict requested bound. Setting the exact expression to $10^{-6}\mathrm{m}$ gives the high-mass thresholdChoose a mass strictly greater than this threshold for a strict inequality; $1.28\times 10^3\mathrm{kg}$ is a suitable rounded choice. Simply rounding the high-frequency estimate down would not guarantee the bound.

The minus solution is the acoustic phonon branch. As $q\to 0$, ${\omega }_{-}$ tends to zero and $U=V$: adjacent atoms move together in a long-wavelength compression. The plus solution is the optical phonon branch, with nonzero zone-centre frequency and $m_{A}U+m_{B}V=0$: the two sublattices move against one another with fixed cell centre of mass. For oppositely charged ions, this relative displacement can couple to an electric field, motivating the optical name.

At the zone boundary the frequencies are $\sqrt{2\alpha /m_{\mathrm{heavy}}}$ and $\sqrt{2\alpha /m_{\mathrm{light}}}$. Different masses produce a gap; equal masses make the two branches meet, recovering a folded monatomic dispersion.

For a hydrogenic donor state, replace the bare electron mass by $m^{\ast }=0.1m_e$ and the Coulomb permittivity by ${ϵ}_r{ϵ}_0$ with ${ϵ}_r=16$. Rescaling the Bohr model gives the positive ground-state binding magnitudeThe radius isusing the Bohr radius $a_0=5.29177\times 10^{-11}\mathrm{m}$. The supplied positive hydrogen “energy” formula should be interpreted as a binding magnitude; the bound level's energy relative to the conduction-band edge is $-E_B$.

At $300\mathrm{K}$, $k_{B}T=25.9\mathrm{meV}$, about five times the binding energy. Thus a substantial fraction, ordinarily most of the shallow donors in the dilute nondegenerate regime, will be ionised. An exact fraction also depends on the Fermi level, donor concentration and degeneracies; comparing $E_B$ with $k_{B}T$ alone does not determine it exactly. The large radius explains why a weakly bound donor extends across many unit cells.

A band effective mass describes the acceleration of an electronic wavepacket in a periodic lattice through the local band curvature. It incorporates the periodic lattice potential into a useful transport parameter without changing the bare electron mass. For a Bloch state in an isolated band,Differentiation givesIn one dimension this becomes $\boxed{m^{\ast }={\hslash }^2/(d^2ϵ/d{k}^2)}$. A locally parabolic isotropic band minimum has $ϵ\simeq {ϵ}_c+{\hslash }^2|\mathbf{k}-{\mathbf{k}}_c{|}^2/(2m^{\ast })$ with positive $m^{\ast }$. A flatter band has a larger effective mass. At a band maximum the electron curvature is negative, so the electron accelerates oppositely to the applied force. Describing the missing electron as a positive-charge electron hole with positive mass is often simpler.

In the scalar relaxation-time approximation, a conduction electron in an electric field reaches ${\mathbf{v}}_d=-e\tau \mathbf{E}/m^{\ast }$. ThusFor a crystal with different directional curvatures, conductivity is a tensor and involves the occupied states and their scattering. Contributions from electrons and holes add: $\sigma =e(n{\mu }_e+p{\mu }_h)$ in an isotropic two-carrier description. A filled band contributes no ordinary longitudinal current because opposite group velocities cancel; a partially filled band or holes permit a current response.

Effective mass therefore connects measured electrical transport to electronic band structure, but carrier density and momentum relaxation time matter as much as curvature. Ionised impurities and phonons alter $\tau$. A smaller mass does not guarantee a higher conductivity if carrier population or scattering changes. Nor are all reported masses identical: density-of-states, cyclotron and conductivity masses weight anisotropic bands differently. The simple semiclassical formula assumes an isolated band, slowly varying weak fields and negligible interband transitions; near degeneracies or strong fields a single constant scalar mass is inadequate.

For a neutral atom, count both the nucleons and the electrons. Protons, neutrons and electrons are all fermions, and exchanging two identical intact atoms produces the sign $(-1{)}^{N_f}$ from exchanging their fermionic constituents. An odd number gives a composite fermion; an even number gives a composite boson, consistently with the Spin-statistics theorem.

A lithium atom has three protons and three electrons. Lithium-6 also has three neutrons, giving $3+3+3=9$ fermionic constituents; lithium-7 has four neutrons, giving $3+4+3=10$. ThereforeThe composite statistics of neutral atoms cannot be inferred from the mass number alone: counting only nucleons would classify the nuclei and give the opposite answers here. The atom's total spin includes its electrons as well as its nucleus.

SetAll are real and $k_0>k_1>k_2>0$. Suppressing a common time quantum phase $e^{-iEt/\hslash }$, chooseThe rightmost region has no left-moving term because no wave is incident from the right. The shifted quantum phase convention makes $F$ the transmitted amplitude at the second interface.

For finite potential steps and equal mass, wavefunction matching at a finite potential step requires continuity of both $\psi$ and ${\partial }_x\psi$. Explicitly,A discontinuous wavefunction would introduce an inadmissible singular kinetic term, while integrating the stationary Schrödinger equation across a finite step shows that the derivative cannot jump. These conditions also keep the probability current continuous, so a finite interface neither creates nor destroys probability. The derivative rule would need modification for a delta-function potential or a different effective mass; neither occurs here.

The one-dimensional time-independent Schrödinger equation isHere $m$ is the particle mass, $V(x)$ its potential energy, $E$ the stationary energy eigenvalue, and $\hslash =h/(2\pi )$ the reduced Planck constant. The spatial wavefunction is $\Psi$; the full stationary state is $\Psi (x)e^{-iEt/\hslash }$. Physical bound states satisfy the boundary conditions and wavefunction normalization $\int |\Psi {|}^{2}dx=1$. The operator on the left is the Hamiltonian operator, the sum of kinetic and potential energies.

Quantum tunnelling is transmission through a finite classically forbidden region, where $E<V(x)$. The stationary Schrödinger equation permits an evanescent wavefunction inside such a barrier rather than forcing it to vanish at its entrance. Matching to propagating states beyond a barrier of finite width gives a nonzero transmitted probability current.

For a rectangular barrier of height $V_0$ and width $a$, with equal exterior potentials and $0<E<V_0$, write $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The exact rectangular-barrier tunnelling probability isFor an opaque barrier its dominant dependence is $e^{-2\kappa a}$. For a slowly varying barrier, the WKB approximation generalizes this to $T\sim \exp [-(2/\hslash ){\int }_{x_1}^{x_2}\sqrt{2m(V(x)-E)}dx]$, apart from matching prefactors. Barrier width, particle mass and energy therefore have a strong influence. Energy is conserved in a static barrier: tunnelling does not require the particle to borrow energy or acquire a classical path with negative kinetic energy.

In alpha decay, a nuclear alpha-particle configuration is confined by the short-range nuclear interaction but separated from the exterior by a Coulomb barrier. Its energy can be positive relative to infinity while still lying below the barrier maximum. Escape by tunnelling produces alpha particles at that conserved energy, with an exponentially small decay probability per nuclear encounter. The pronounced dependence of half-life on alpha energy is a characteristic consequence of the barrier exponent, unlike a simple classical escape threshold.

In scanning tunneling microscopy, a conducting tip is brought within a nanometer-scale vacuum gap of a conducting surface and biased weakly. Electron tunnelling gives a current that depends approximately exponentially on separation, $I\propto V{e}^{-2\kappa s}$ in a simple low-bias model. Feedback adjusts the tip height to keep current fixed while scanning, producing a surface map sensitive to both geometry and electronic states. Finite current through the non-contact gap, and its steep dependence on separation, are the direct tunnelling signatures; current is not a direct measurement of topographic height alone.

A Josephson junction provides a collective superconducting example. Two superconductors separated by a thin barrier can carry a coherent pair current $I=I_c\sin \phi$ even at zero voltage. At constant voltage the AC Josephson effect gives $d\phi /dt=2eV/\hslash$ and oscillation frequency $2eV/h$. The coupled condensate quantum phases and pair transfer distinguish this from independent normal-electron transmission. These examples involve different microscopic barriers, but all require quantum coupling across a region that suppresses classical passage. Comparing weak-temperature tunnelling behavior with thermally activated escape helps distinguish the mechanisms experimentally.

Take nearest-neighbour separation $d$, a common spring constant $K$, and one scalar displacement per atom. For a monatomic chain, the harmonic approximation near a potential minimum, Newton's second law and a normal mode giveThe monatomic-chain dispersion relation has one acoustic phonon branch, with $\omega \simeq d\sqrt{K/m}|k|$ near the origin. The first Brillouin zone ends at $\boxed{k=\pm \pi /d}$, where $\omega =2\sqrt{K/m}$ and the group velocity vanishes.

For alternating masses $m,M$, choose a two-atom cell of length $2d$. The coupled normal modes give the diatomic-chain dispersion relation with equal springs,The first Brillouin zone now ends at $\boxed{k=\pm \pi /(2d)}$. The minus branch is the acoustic phonon branch; its two atoms move together at long wavelength. The plus branch is the optical phonon branch; at $k=0$ the atoms move oppositely with a stationary centre of mass and ${\omega }_{+}^2=2K(1/m+1/M)$. For $M>m$ the edge frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, with a gap between them.

If $M=m$, the two branches reduce to $2\sqrt{K/m}|\sin (kd/2)|$ and $2\sqrt{K/m}|\cos (kd/2)|$, touching at $\sqrt{2K/m}$. This is zone folding in an equal-mass diatomic chain: identical sites actually have a primitive unit cell of length $d$. A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour spring constants.

The symbol $\rho$ here is linear mass density, with units $\mathrm{kg}{\mathrm{m}}^{-1}$. A short element $[x,x+\delta x]$ has transverse forcebecause small slopes give $\sin \theta \simeq \tan \theta ={\Psi }_x$. Its mass is $\rho \delta x$. Applying Newton's second law, dividing by $\delta x$, and taking the limit gives the wave equation on a string,The derivation assumes constant positive tension, small transverse displacements and slopes, a flexible string with negligible bending stiffness, and negligible damping. Variations of tension due to the motion enter beyond the linear approximation.

At an ordinary point or a massless join, the string is continuous, soWith no concentrated transverse force or point mass, transverse force balance requiresFor the common tension in this problem, the slope $\partial \Psi /\partial x$ is therefore continuous as well. A jump in linear mass density changes wave speed, but introduces no finite inertial force at a zero-length join. A point mass, external support force, or unequal tensions would require modified boundary conditions.

A Bloch state in a crystal experiences a periodic lattice potential. Its band dispersion relation $ϵ(k)$ encodes the lattice response, so an effective mass allows a wavepacket's acceleration under a weak applied force to be described without explicitly following the microscopic periodic forces.

In one dimension the wavepacket group velocity and semiclassical equation for crystal momentum areDifferentiating the first equation with respect to time givesDefining $F=m^{\ast }a$ yieldsThis band effective mass is controlled by band curvature, not the slope. A band minimum has positive $m^{\ast }$; a band maximum has negative electron $m^{\ast }$ and is often more conveniently described with positive-mass electron holes. In three dimensions the inverse effective mass is generally a tensor, $(m^{\ast -1}{)}_{ij}={\hslash }^{-2}{\partial }_{k_i}{\partial }_{k_j}ϵ$. The scalar formula assumes motion along a specified direction in an isolated band, weak fields and negligible interband transitions; the bare electron mass has not physically changed.

The first two probability densities areBoth vanish at the walls. The ground state has one maximum at $L/2$ and no interior node. The first excited state has a node at $L/2$ and equal maxima at $L/4$ and $3L/4$.

For a finite square well whose exterior potential is $V_0>E_n$, the exterior wavefunction decays as $e^{-\kappa d}$, where $\kappa =\sqrt{2m(V_0-E_n)}/\hslash$ and $d$ is distance from the relevant edge. Wavefunction matching at a finite potential step makes the wavefunction and, for constant mass, its first derivative continuous. Thus the probability density is nonzero at the edges and has evanescent tails outside, the interior wavelength increases, and the energy levels are lower than for infinite walls at the same positions. Reflection symmetry still gives an even ground state and odd first excited state about the center, when both are bound. A shallow well may not support that excited bound state at all.

Treat sodium as monovalent, with one mobile electron per atom. Its mass density is ${\rho }_m=0.97\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}=970\mathrm{kg}{\mathrm{m}}^{-3}$, and its molar mass is $M=0.023\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Using the Avogadro constant,The electron mobility $\mu =e\tau /m_e$ relates the magnitude of drift velocity to electric field. The Drude model then givesUsing $m_e=9.109\times 10^{-31}\mathrm{kg}$,The inferred time is the momentum relaxation time. Identifying it with the mean time between collisions is the intended simple isotropic Drude model assumption; small-angle scattering need not relax directed momentum in a single event.

Here $x(t)$ is displacement from equilibrium, in metres; $m>0$ is the oscillating mass, in kilograms; $b\ge 0$ is the coefficient of linear viscous damping, in $\mathrm{kg}{\mathrm{s}}^{-1}$; and $k>0$ is the spring constant, in $\mathrm{N}{\mathrm{m}}^{-1}$. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ are respectively the inertial term, the opposing velocity-dependent damping term and the opposing displacement-dependent restoring term when moved to the force side of Newton's second law.

$F_0$ is a possibly complex driving-force complex amplitude, measured in newtons. $\omega$ is the driving angular frequency in $\mathrm{rad}{\mathrm{s}}^{-1}$, and taking the real part of $F_0{e}^{i\omega t}$ gives the physical sinusoidal force. Its frequency in hertz is $f=\omega /(2\pi )$. The natural undamped angular frequency and damping rate areThis damped harmonic oscillator model assumes a linear restoring force and a linear, instantaneous viscous damping law.